A capacitor `4 muF` charged to `50 V` is connected to another capacitor of `2 muF` charged to `100 V` with plates of like charges connected together. The total energy before and after connection in multiples of `(10^(-2) J)` is

Options

A. `1.5` and `1.33`
B. `1.33` and `1.5`
C. `3.0` and `2.67`
D. `2.67` and `3.0`

Explanation

To solve the problem, we need to calculate the total energy stored in both capacitors before and after they are connected. Let's break it down step by step.

### Step 1: Calculate the initial energy stored in each capacitor.

The energy (U) stored in a capacitor can be calculated using the formula:
\[ U = \frac{1}{2} C V^2 \]
where \(C\) is the capacitance and \(V\) is the voltage.

**For the 4 µF capacitor charged to 50 V:**
\[ U_1 = \frac{1}{2} \times 4 \times 10^{-6} \, \text{F} \times (50 \, \text{V})^2 \]
\[ U_1 = \frac{1}{2} \times 4 \times 10^{-6} \times 2500 \]
\[ U_1 = \frac{1}{2} \times 10^{-2} \, \text{J} = 5 \times 10^{-3} \, \text{J} \]

**For the 2 µF capacitor charged to 100 V:**
\[ U_2 = \frac{1}{2} \times 2 \times 10^{-6} \, \text{F} \times (100 \, \text{V})^2 \]
\[ U_2 = \frac{1}{2} \times 2 \times 10^{-6} \times 10000 \]
\[ U_2 = \frac{1}{2} \times 2 \times 10^{-2} \, \text{J} = 1 \times 10^{-2} \, \text{J} \]

### Step 2: Calculate the total initial energy.

Now, we can find the total initial energy:
\[ U_{\text{total initial}} = U_1 + U_2 \]
\[ U_{\text{total initial}} = 5 \times 10^{-3} \, \text{J} + 1 \times 10^{-2} \, \text{J} \]
\[ U_{\text{total initial}} = 1.5 \times 10^{-2} \, \text{J} \]

### Step 3: Calculate the final voltage after connection.

When the capacitors are connected with like charges together, the total charge will redistribute. The total charge before connection is:
\[ Q_1 = C_1 V_1 = 4 \times 10^{-6} \times 50 = 2 \times 10^{-4} \, \text{C} \]
\[ Q_2 = C_2 V_2 = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4} \, \text{C} \]

The total charge \(Q_{\text{total}} = Q_1 + Q_2 = 2 \times 10^{-4} + 2 \times 10^{-4} = 4 \times 10^{-4} \, \text{C}\).

The total capacitance when connected in parallel is:
\[ C_{\text{total}} = C_1 + C_2 = 4 \times 10^{-6} + 2 \times 10^{-6} = 6 \times 10^{-6} \, \text{F} \]

Now, we can find the final voltage \(V_f\) using:
\[ V_f = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{4 \times 10^{-4}}{6 \times 10^{-6}} = \frac{4}{6} \times 10^2 = \frac{2}{3} \times 100 \approx 66.67 \, \text{V} \]

### Step 4: Calculate the final energy stored in the combined capacitor.

Using the final voltage, we can find the energy stored in the combined capacitor:
\[ U_{\text{final}} = \frac{1}{2} C_{\text{total}} V_f^2 \]
\[ U_{\text{final}} = \frac{1}{2} \times 6 \times 10^{-6} \times (66.67)^2 \]
\[ U_{\text{final}} = \frac{1}{2} \times 6 \times 10^{-6} \times 4444.44 \approx 1.33 \times 10^{-2} \, \text{J} \]

### Step 5: Calculate the total energy before and after connection in multiples of \(10^{-2} J\).

**Total initial energy:**
\[ U_{\text{total initial}} = 1.5 \times 10^{-2} \, \text{J} \]

**Total final energy:**
\[ U_{\text{final}} \approx 1.33 \times 10^{-2} \, \text{J} \]

### Final Answer:
The total energy before connection is \(1.5 \times 10^{-2} J\) and after connection is approximately \(1.33 \times 10^{-2} J\).